Answer:
[tex]2.11\ * 10^{-2}[/tex] is the correct answer to the given question.
Explanation:
Given k=6.40 x 10-3 min-1.
According to the first order reaction .
The concentration of time can be written as
[tex][\ A\ ]\ = \ [\ A_{0}\ ] * e \ ^\ {-kt}[/tex]
Here [tex][\ A\ ]_{0}[/tex] = Initial concentration.
So [tex][\ A\ ]_{0}= 0.0314 M[/tex]
Putting this value into the above equation.
[tex]0.0314 \ *\ e^{6.40 x 10^{-3} \ * \ 62.0 }[/tex]
=0.211 M
This can be written as
[tex]=\ 2.11 *\ 10^{-2}[/tex]
